Q.Prove the following: $\sec840°\cdot\cot(-945°)+\sin600°\cdot\tan(-690°)=\dfrac{3}{2}$

Step 1: $\sec840°=\sec(2\times360°+120°)=\sec120°=\sec(90°+30°)=-\text{cosec }30°=-2$.

Step 2: $\cot(-945°)=-\cot945°=-\cot(2\times360°+225°)=-\cot225°=-\cot(180°+45°)=-\cot45°=-1$.

Step 3: $\sin600°=\sin(360°+240°)=\sin240°=\sin(180°+60°)=-\sin60°=-\dfrac{\sqrt3}{2}$.

Step 4: $\tan(-690°)=-\tan690°=-\tan(2\times360°-30°)=-(-\tan30°)=\dfrac{1}{\sqrt3}$. …

Reduce each of the four ratios independently to a standard angle, …

• Losing track of the double negative in $\cot(-945°)=-\cot945°$ …